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The distance between the lenses in a compound microscope is \(18 \mathrm{cm} .\) The focal length of the objective is \(1.5 \mathrm{cm} .\) If the microscope is to provide an angular magnification of -83 when used by a person with a normal near point \((25 \mathrm{cm}\) from the eye), what must be the focal length of the eyepiece?

Short Answer

Expert verified
Focal length of the eyepiece must be approximately \(5.083\, \mathrm{cm}.\)

Step by step solution

01

Identify Given Parameters

We have the following information: distance between lenses (tube length) \(L = 18 \mathrm{cm}\); focal length of the objective \(f_o = 1.5 \mathrm{cm}\); desired angular magnification \(M = -83\); and the normal near point \(N = 25 \mathrm{cm}\). We need to find the focal length of the eyepiece \(f_e\).
02

Understand Relationship Between Elements

For a compound microscope, the magnification \(M\) is given by the formula:\[ M = -\left( \frac{L}{f_o} \right) \left( 1 + \frac{N}{f_e} \right) \]We will solve this equation for \(f_e\).
03

Substitute Known Values into the Formula

Replace the known values into the equation:\[ -83 = -\left( \frac{18}{1.5} \right) \left( 1 + \frac{25}{f_e} \right) \]
04

Simplify the Magnification Equation

Calculate \( \frac{18}{1.5} \), resulting in \(12\). Thus, the equation becomes:\[ -83 = -12 \left( 1 + \frac{25}{f_e} \right) \]
05

Resolve Magnification Equation to Isolate \(f_e\)

Divide both sides by \(-12\) and simplify:\[ \frac{83}{12} = 1 + \frac{25}{f_e} \]
06

Solve for Eyepiece Focal Length

Subtract 1 from both sides of the equation:\[ \frac{83}{12} - 1 = \frac{25}{f_e} \]Calculate \( \frac{83}{12} - 1 \approx 5.916667-1 = 4.916667\).Thus, \( \frac{25}{f_e} = 4.916667 \).
07

Calculate Focal Length \(f_e\)

Rearrange to find \(f_e\):\[ f_e = \frac{25}{4.916667} \]Calculate \(f_e \approx 5.083 \mathrm{cm}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Compound Microscope
A compound microscope is a powerful optical device, widely used in scientific research and classrooms to observe small objects in detail. It uses two lenses to magnify the image of the object being studied.

The first lens, known as the objective lens, is positioned close to the object. It creates an enlarged image of the object within the body of the microscope. The eyepiece, which is the second lens, is located near the viewer's eye and magnifies the image created by the objective lens.

Together, these lenses significantly enhance the size of tiny objects, making them visible to the naked eye. This process of using two lenses provides greater magnification than a simple microscope, which only uses a single lens. Compound microscopes are essential for examining details that are too small to be seen with the naked eye.
Focal Length Calculation
Understanding focal length is crucial in the use of lenses, and it plays a significant role in microscopes. Focal length refers to the distance between the lens and the image sensor when the object is in focus. In our case, we are interested in calculating the focal length of the eyepiece of a compound microscope.

To find the focal length of the eyepiece ( f_e), we employ the magnification formula relating the objective lens and the eyepiece: \[ M = - \left( \frac{L}{f_o} \right) \left( 1 + \frac{N}{f_e} \right) \]where L is the distance between the lenses, f_o is the focal length of the objective lens, N is the normal near point, and M is the magnification required.

Rearranging and solving this equation allows us to determine the necessary focal length of the eyepiece so that the microscope provides the desired magnifying power. Proper calculation of focal lengths ensures that microscopes are accurately set up for viewing samples at the correct scale.
Angular Magnification
Angular magnification is a measure of how much larger a lens or optical system can make an object appear through a specific angle change as observed by the eye. This concept is central to understanding how telescopes and microscopes function.

In microscopes, angular magnification is not only determined by the lenses' focal lengths but also by how these lenses are positioned relative to each other. It is defined by the relationship: \[ M = - \left( \frac{L}{f_o} \right) \left( 1 + \frac{N}{f_e} \right) \]Where M represents angular magnification, L the distance between lenses, f_o the objective lens focal length, N the normal near point, and f_e the eyepiece focal length.

Adjusting these variables in the microscope allows users to optimize the angular magnification to suit specific observing needs. The negative sign indicates that the image produced is inverted—common in compound microscopes. Through this understanding, microscopes can be finely tuned to achieve detailed and magnified observations of tiny specimens.

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Most popular questions from this chapter

Two converging lenses are separated by \(24.00 \mathrm{cm} .\) The focal length of each lens is \(12.00 \mathrm{cm} .\) An object is placed \(36.00 \mathrm{cm}\) to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.

An engraver uses a magnifying glass \((f=9.50 \mathrm{cm})\) to examine some work, as in Figure \(26.39 b .\) The image he sees is located \(25.0 \mathrm{cm}\) from his eye, which is his near point. (a) What is the distance between the work and the magnifying glass? (b) What is the angular magnification of the magnifying glass?

A tourist takes a picture of a mountain \(14 \mathrm{km}\) away using a camera that has a lens with a focal length of \(50 \mathrm{mm}\). She then takes a second picture when she is only \(5.0 \mathrm{km}\) away. What is the ratio of the height of the mountain's image on the camera's image sensor for the second picture to its height on the image sensor for the first picture?

An object is located \(30.0 \mathrm{cm}\) to the left of a converging lens whose focal length is \(50.0 \mathrm{cm} .\) (a) Draw a ray diagram to scale and from it determine the image distance and the magnification. (b) Use the thin-lens and magnification equations to verify your answers to part (a).

Emergency Replacement Glasses. One of your team members lost her glasses in a river. She is nearsighted and cannot see long distances without them. You are on an expedition to map a remote area in southern Argentina, and her long- distance vision is crucial to her role in the group. You can request a new pair of glasses for her that will be delivered with the next airdrop of supplies, but she does not know her lens prescription. (a) Does she need a converging or diverging lens? (b) You do a simple eye test and estimate her far point to be \(623.0 \mathrm{cm}\) from her eyes. Assuming she will wear her glasses \(2.0 \mathrm{cm}\) in front of her eyes, what should be the focal length of her new lenses? (c) What should be the refractive power of her new lenses (in diopters)?

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